Đáp án:
c. C=0
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a.A = a\left( {\frac{1}{2} - \frac{2}{3} + \frac{3}{4}} \right) = a.\left( {\frac{{6 - 2.4 + 3.3}}{{12}}} \right)}\\
\begin{array}{l}
= a.\left( {\frac{{ - 2 + 9}}{{12}}} \right) = a.\frac{7}{{12}}\\
Thay:a = - \frac{6}{5}\\
\to A = - \frac{6}{5}.\frac{7}{{12}} = {\rm{\;}} - \frac{7}{{10}}
\end{array}\\
{b.B = b\left( { - \frac{1}{6} + \frac{4}{3} - \frac{1}{2}} \right) = b\left( {\frac{{ - 1 + 8 - 3}}{6}} \right)}\\
\begin{array}{l}
= b\left( {\frac{4}{6}} \right) = b.\frac{2}{3}\\
Thay:b = - \frac{3}{7}\\
\to B = - \frac{3}{7}.\frac{2}{3} = {\rm{\;}} - \frac{2}{7}
\end{array}\\
{c.C = c\left( {\frac{5}{4} + \frac{1}{6} - \frac{{17}}{{12}}} \right) = c.\left( {\frac{{15 + 2 - 17}}{{12}}} \right)}\\
\begin{array}{l}
= c.0\\
Thay:c = \frac{{2013}}{{2014}}\\
\to C = \frac{{2013}}{{2014}}.0 = 0
\end{array}
\end{array}\)
